Problem: Let $f(x)=x^3-2 x^2-4 x+8 $. For what value of $x$ does $f$ have a relative minimum ? Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $-2$ (Choice C) C $2$ (Choice D) D $-\dfrac{2}{3}$
Explanation: We can find the relative extrema (i.e. minima and maxima) of $f$ by looking for the intervals where its derivative $f'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $f$ is $f'(x)=(3x+2)(x-2)$. $f'(x)=0$ for $x=-\dfrac23,2$. Since $f'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-\dfrac23$ and $x=2$. $f$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $x<-\frac{2}{3}$ $-\frac{2}{3}<x<2$ $-\frac{2}{3}$ $x>2$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $x<-\dfrac{2}{3}$ $x=-1$ $f'(-1)=3>0$ $f$ is increasing $\nearrow$ $-\dfrac{2}{3}<x<2$ $x=0$ $f'\left(0\right)=-4<0$ $f$ is decreasing $\searrow$ $x>2$ $x=3$ $f'(3)=11>0$ $f$ is increasing $\nearrow$ Now let's look at the critical points: $x$ Before After Verdict $-\dfrac{2}{3}$ $\nearrow$ $\searrow$ Maximum $2$ $\searrow$ $\nearrow$ Minimum Now we can see that $f$ has a relative minimum at $x=2$.